Differential Equations Quiz: Common Setup Pitfalls

What this quiz covers

This quiz focuses on Common Setup Pitfalls, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $dP/dt = 0.2P(1 - P/10000) + 300, \quad P(0)=2000$
2. $dP/dt = 0.2P(1 - P/2000) - 300, \quad P(0)=10000$
3. $dP/dt = 0.2(1 - P/10000) - 300, \quad P(0)=2000$
4. $dP/dt = 0.2P(1 - P/10000) - 300, \quad P(0)=2000$ (correct answer)

Explanation: The standard logistic model with intrinsic growth rate $r$ and carrying capacity $M$ is $dP/dt = rP(1 - P/M)$. Here, $r=0.2$ and $M=10000$. Harvesting removes fish from the population, so the harvesting rate is subtracted from the growth rate. The harvesting is at a constant rate of 300 fish per year. Therefore, the differential equation is $dP/dt = 0.2P(1 - P/10000) - 300$. The initial population is given as 2,000, so $P(0)=2000$.

Question 2

1. $y(0) = 0.1, \quad y'(0) = 0.5$
2. $y(0) = -0.1, \quad y'(0) = -0.5$
3. $y(0) = 0.1, \quad y'(0) = -0.5$ (correct answer)
4. $y(0) = -0.1, \quad y'(0) = 0.5$

Explanation: The displacement $y(t)$ is measured from the equilibrium position, with the downward direction as positive. 'Pulled down 0.1 meters' means the initial position is 0.1 m in the positive direction. Therefore, $y(0) = 0.1$. The initial velocity $y'(0)$ is the speed and direction at $t=0$. 'Pushed upwards with a speed of 0.5 m/s' means the initial velocity is in the negative direction. Therefore, $y'(0) = -0.5$. The correct initial conditions are $y(0) = 0.1$ and $y'(0) = -0.5$.

Question 3

1. $500 dv/dt = 500(9.8) - 0.2v, \quad v(0)=0$
2. $0.5 dv/dt = 0.5(9.8) + 0.2v, \quad v(0)=0$
3. $0.5 dv/dt = -0.5(9.8) - 0.2v, \quad v(0)=0$
4. $0.5 dv/dt = 0.5(9.8) - 0.2v, \quad v(0)=0$ (correct answer)

Explanation: By Newton's second law, $m(dv/dt) = F_{\text{net}} = F_{\text{gravity}} + F_{\text{resistance}}$. A critical first step is to ensure consistent units. The mass is given as 500 grams, which must be converted to kilograms to be compatible with Newtons and m/s$^2$: $m = 0.5$ kg. The force of gravity acts in the positive (downward) direction, so $F_{\text{gravity}} = mg = 0.5(9.8)$ N. Air resistance opposes motion, so it acts in the negative (upward) direction: $F_{\text{resistance}} = -0.2v$. The object is 'dropped from rest', so the initial velocity is $v(0)=0$. Combining these gives the initial value problem: $0.5 dv/dt = 0.5(9.8) - 0.2v$, with $v(0)=0$.

Question 4

1. $0.5q'' + 10q' + 100q = 12, \quad q(0)=2, \ q'(0)=0$ (correct answer)
2. $0.5q'' + 10q' + 100q = 12, \quad q(0)=0, \ q'(0)=2$
3. $0.5q'' + 10q' + 0.01q = 12, \quad q(0)=2, \ q'(0)=0$
4. $0.5q'' + 10q' + 100q = 0, \quad q(0)=2, \ q'(0)=0$

Explanation: The differential equation for an RLC circuit is given by Kirchhoff's voltage law: $Lq'' + Rq' + \frac{1}{C}q = E(t)$. Substituting the given values: $L=0.5$, $R=10$, $C=0.01$ (so $1/C = 100$), and $E(t)=12$. This gives the equation $0.5q'' + 10q' + 100q = 12$. The initial conditions are derived from the problem statement. 'The initial charge on the capacitor is 2 C' means $q(0)=2$. 'There is no initial current' means $I(0)=0$. Since current $I$ is the rate of change of charge, $I(t) = q'(t)$, so $q'(0)=0$. This yields the IVP in option A.

Question 5

1. $dP/dt = (\beta + \delta)P$
2. $dP/dt = (\beta - \delta)P$ (correct answer)
3. $dP/dt = \beta P - \delta$
4. $dP/dt = (\delta - \beta)P$

Explanation: When modeling population dynamics, you need to carefully account for all factors affecting the rate of change and determine whether each factor increases or decreases the population. Let's break down this problem systematically. The population experiences two effects:

• Birth rate: proportional to current population size with constant $\beta$, contributing $+\beta P$ to the rate of change
• Death rate from predation: proportional to current population size with constant $\delta$, contributing $-\delta P$ to the rate of change

Question 6

1. $dS/dt = 12 - S/50, \quad S(0) = 0$
2. $dS/dt = 12 - S/50, \quad S(5) = 0$ (correct answer)
3. $dS/dt = 3 - S/200, \quad S(5) = 0$
4. $dS/dt = 12 - S/50, \quad S(5) = 60$

Explanation: When modeling mixing problems with differential equations, you need to set up the rate equation $\frac{dS}{dt} = \text{rate in} - \text{rate out}$ and carefully identify the initial condition at the specific time period you're analyzing. For $t \geq 5$, salt enters at 4 gal/min with concentration 3 lb/gal, so the rate in is $4 \times 3 = 12$ lb/min. Since the tank volume stays constant at 200 gallons (inflow equals outflow), the rate out is $\frac{S(t)}{200} \times 4 = \frac{S}{50}$ lb/min. This gives us $\frac{dS}{dt} = 12 - \frac{S}{50}$. The key insight is determining the initial condition. At $t = 5$, the tank has contained only pure water for 5 minutes, so $S(5) = 0$ pounds of salt. This is when we start tracking salt accumulation. Answer B correctly captures both the differential equation and initial condition: $\frac{dS}{dt} = 12 - \frac{S}{50}, S(5) = 0$. Answer A uses the wrong initial condition $S(0) = 0$, but we're modeling the system starting at $t = 5$, not $t = 0$. Answer C incorrectly calculates the rate out as $\frac{S}{200}$ instead of $\frac{4S}{200} = \frac{S}{50}$, missing the outflow rate multiplication. Answer D uses $S(5) = 60$, but there's no salt in the tank at $t = 5$ since only pure water entered initially. Study tip: In mixing problems, always multiply concentration by flow rate for both inflow and outflow terms, and set your initial condition at the time when the scenario you're modeling begins.

Question 7

1. $dx/dt = k_1(a_0 - x) + k_2 x$
2. $dx/dt = k_2 x - k_1(a_0 - x)$
3. $dx/dt = k_1(a_0 - x) - k_2 x$ (correct answer)
4. $dx/dt = k_1 a_0 - k_2 x$

Explanation: The rate of change of the concentration of B, $dx/dt$, is equal to the rate of its formation minus the rate of its consumption. B is formed by the forward reaction $A \to B$, so the rate of formation is $k_1[A]$. B is consumed by the reverse reaction $B \to A$, so the rate of consumption is $k_2[B]$. This gives $dx/dt = k_1[A] - k_2[B]$. We are given $[B] = x(t)$. By stoichiometry, the concentration of A at time $t$ is its initial concentration minus the amount that has reacted to form B, so $[A] = a_0 - x(t)$. Substituting these into the rate equation gives $dx/dt = k_1(a_0 - x) - k_2 x$.

Question 8

1. $dT/dt = k(T - 40)$, $T(0) = 5$
2. $dT/dt = -k(40 - T)$, $T(0) = 5$
3. $dT/dt = k(40 - T)$, $T(0) = 40$
4. $dT/dt = k(40 - T)$, $T(0) = 5$ (correct answer)

Explanation: The rate of change, $dT/dt$, is proportional to the difference between the ambient temperature ($40^\circ\text{C}$) and the object's temperature ($T$). So, $dT/dt = k(40 - T)$ for some constant $k$. Since the probe is warming up ($T(0)=5$ and the bath is at $40$), the temperature $T$ must increase, meaning $dT/dt$ must be positive when $T < 40$. The term $(40 - T)$ is positive for $T < 40$, so the constant $k$ must be positive. The initial condition is explicitly given as $T(0) = 5$. Thus, the correct model is $dT/dt = k(40 - T)$ with $T(0) = 5$.

Question 9

1. $dh/dt = -k\sqrt{h}$
2. $dh/dt = -\frac{k}{4\pi}\sqrt{h}$ (correct answer)
3. $dh/dt = \frac{k}{4\pi}\sqrt{h}$
4. $dh/dt = -4\pi k\sqrt{h}$

Explanation: We are given the rate of change of volume, $dV/dt = -k\sqrt{h}$. We need to find the rate of change of height, $dh/dt$. For a cylinder, the volume is $V = A \cdot h$, where $A$ is the constant cross-sectional area. Here, the radius is $r=2$ m, so the area is $A = \pi r^2 = 4\pi$ m$^2$. Thus, $V = 4\pi h$. We can relate the rates using the chain rule: $dV/dt = \frac{dV}{dh} \frac{dh}{dt}$. From $V = 4\pi h$, we have $dV/dh = 4\pi$. Substituting this into the chain rule expression gives $dV/dt = 4\pi \frac{dh}{dt}$. Now we equate the two expressions for $dV/dt$: $4\pi \frac{dh}{dt} = -k\sqrt{h}$. Solving for $dh/dt$ yields $dh/dt = -\frac{k}{4\pi}\sqrt{h}$.

Question 10

1. $dP/dt = 0.06P + K, \quad P(0)=30000$
2. $dP/dt = K - 0.06P, \quad P(0)=30000$
3. $dP/dt = 0.06P - K, \quad P(0)=30000$ (correct answer)
4. $dP/dt = 0.005P - K, \quad P(0)=30000$

Explanation: The rate of change of the loan balance, $dP/dt$, is the rate at which the balance grows due to interest minus the rate at which it is reduced by payments. The interest accrues at a rate of $rP$, where $r=0.06$, so this term is $0.06P$. The payments reduce the loan, so they are subtracted at a rate of $K$ per year. This leads to the differential equation $dP/dt = 0.06P - K$. The initial loan amount is $30,000, so the initial condition is$P(0) = 30000$.

Question 11

1. $dA/dt = 12 - \frac{\ln(2)}{3}A$ (correct answer)
2. $dA/dt = 12 + \frac{\ln(2)}{3}A$
3. $dA/dt = 12 - 3A$
4. $dA/dt = -12 - \frac{\ln(2)}{3}A$

Explanation: The rate of change of the amount, $dA/dt$, is the rate of creation minus the rate of decay. The rate of creation is given as 12 g/hr. Radioactive decay occurs at a rate proportional to the amount present, so the rate of decay is $kA$ for some decay constant $k>0$. Thus, the model is $dA/dt = 12 - kA$. The decay constant $k$ is related to the half-life $T_{1/2}$ by the formula $T_{1/2} = \ln(2)/k$. Given $T_{1/2} = 3$ hours, we can solve for $k$: $3 = \ln(2)/k \implies k = \ln(2)/3$. Substituting this value of $k$ into the model gives $dA/dt = 12 - \frac{\ln(2)}{3}A$.

Question 12

1. $dA/dt = 0.5 - \frac{5A}{400}$
2. $dA/dt = 0.5 - \frac{5A}{1000}$
3. $dA/dt = 0.5 - \frac{5A}{400 + 5t}$ (correct answer)
4. $dA/dt = 0.5 + \frac{5A}{400 + 5t}$

Explanation: The rate of change of salt is $dA/dt = (\text{rate in}) - (\text{rate out})$. The rate of salt flowing in is (concentration in) $\times$ (flow rate in) = $0.05 \text{ kg/L} \times 10 \text{ L/min} = 0.5$ kg/min. The volume of the solution in the tank is not constant; it changes at a rate of $(10 - 5) = 5$ L/min. The initial volume is 400 L, so $V(t) = 400 + 5t$. The concentration of salt in the tank at time $t$ is $C(t) = A(t)/V(t) = A(t)/(400 + 5t)$. The rate of salt flowing out is (concentration out) $\times$ (flow rate out) = $(A(t)/(400 + 5t)) \times 5 \text{ L/min}$. Therefore, the differential equation is $dA/dt = 0.5 - \frac{5A}{400+5t}$.